Consider the circuit shown : The ammeter reads 0.9 A. Value of R is
The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A.
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.
According to Ampere’s law, magnetic fields are related to the electric current that is produced in them. This law specifies that the magnetic field is associated with a given current or vice-versa, provided that the electric field doesn’t change with time.
Ampere’s circuital law can be written as the line integral of the magnetic field surrounding the closed loop which is equal to the number of times the algebraic sum of currents passing through the loop.
According to the second equation, if the magnetic field is integrated along the blue path, then it is equal to the current enclosed, I.
The magnetic field doesn’t vary at a distance r because of symmetry. The path length (in blue) in figure 1 has to be equal to the circumference of a circle,2πr.